Question

4. Six years hence a man's age will be three times the age of his son and three years ago he
was nine times as old as his son. Find their present ages​

Answer 1

Solution :-

Let the ages of a man and his son be x years and y years respectively

Case I : Six years hence a man's age will be three times the age of his son.

=> x + 6 = 3(y + 6)

=> x = 3y + 18 - 6

=> x = 3y + 12 _______(i)

Case II : Three years ago he was nine times as old as his son.

=> x - 3 = 9(y - 3)

=> 3y + 12 - 3 = 9y - 27 [from equation (i)]

=> 6y = 36

=> y = 36/6 = 6

Putting the value of y in equation (i) we get,

=> x = 3 × 6 + 12 = 30

Hence,

The ages of a man and his son are 30 years and 6 years respectively .

Answer 2

Six years hence man's age will be 3 times the age of his son.

Let Age of Son after 6 years=x years

Let Age of Man After 6 years=3x years

Present age of Son=(x-6) years

Present age of Man=(3x-6) years

3 years Ago

Son age was=(x-6-3)=(x-9) years

Father Age was=(3x-6-3)=(3x-9) years

Now 3 years ago Father Age (3x-9) was 9 times his son age 3 years ago(x-9)

Means

Father Age 3 years ago=9(son age 3 years ago)

3x-9=9(x-9)

3x-9=9x-81

3x-9x=-81+9

-6x=-72

x=-72÷-6

x=12

Present age of Son=(x-6)=12-6=6 years

Present age of Father=(3x-6)=3×12-6=30 years

Son=6 years

Father=30 years